关于有趣常数g的数学猜想问询：对数近似式的渐近行为

Let's break down your conjecture and the asymptotic behavior you've observed with concrete algebraic simplification and reasoning.

First, let's tackle the sum term in your expression—it's a geometric series, so we can compute it exactly using the geometric sum formula:
$$
\sum_{i=0}^{n-1} e^{\pi+i} = e^\pi \sum_{i=0}^{n-1} e^i = e^\pi \cdot \frac{e^n - 1}{e - 1}
$$

Substitute this back into the expression inside the logarithm to simplify the core of your conjecture:
$$
e^{\pi+n} - \sum_{i=0}^{n-1} e^{\pi+i} = e^{\pi+n} - e^\pi \cdot \frac{e^n - 1}{e - 1}
$$

Factor out $e^\pi$ to clean up the expression further:
$$
= e^\pi \left( e^n - \frac{e^n - 1}{e - 1} \right) = e^\pi \cdot \frac{(e-1)e^n - e^n + 1}{e-1} = e^\pi \cdot \frac{(e-2)e^n + 1}{e-1}
$$

Now take the natural logarithm of this result:
$$
\ln\left( e^{\pi+n} - \sum_{i=0}^{n-1} e^{\pi+i} \right) = \ln\left( e^\pi \cdot \frac{(e-2)e^n + 1}{e-1} \right) = \pi + \ln\left( \frac{(e-2)e^n + 1}{e-1} \right)
$$

Explaining the Asymptotic Behavior #

When $n \in \mathbb{N}$ grows large, the constant term $1$ in the numerator becomes negligible compared to the exponentially growing $(e-2)e^n$. We can approximate the logarithm term as:
$$
\ln\left( \frac{(e-2)e^n + 1}{e-1} \right) \approx \ln\left( \frac{(e-2)e^n}{e-1} \right) = n + \ln\left( \frac{e-2}{e-1} \right)
$$

Plug this back into your original left-hand expression:
$$
\ln\left( e^{\pi+n} - \sum_{i=0}^{n-1} e^{\pi+i} \right) - \frac{\pi}{\sqrt{2}} \approx \pi + n + \ln\left( \frac{e-2}{e-1} \right) - \frac{\pi}{\sqrt{2}}
$$

Group the constant terms to get the asymptotic form you noticed:
$$
\sim n + \underbrace{\pi\left(1 - \frac{1}{\sqrt{2}}\right) + \ln\left( \frac{e-2}{e-1} \right)}_{k}
$$

Why the Approximation Improves Then Levels Off #

• For small $n$, the $1$ term in $\frac{(e-2)e^n + 1}{e-1}$ isn't negligible. As $n$ increases, the exponential $e^n$ dominates more and more, making the gap between the left-hand side and $n$ shrink.
• Once $n$ is large enough that $1$ is effectively zero compared to $(e-2)e^n$, the approximation stops getting better. The difference stabilizes to the constant $k$, hence the asymptotic relation $\sim n + k$.

If you want a numerical estimate of $k$, plug in approximate values for constants:

• $\pi \approx 3.1416$, $\sqrt{2} \approx 1.4142$, $e \approx 2.7183$
• Calculating each part:
  • $\pi\left(1 - \frac{1}{\sqrt{2}}\right) \approx 3.1416 \times 0.2929 \approx 0.920$
  • $\ln\left( \frac{e-2}{e-1} \right) \approx \ln(0.7183) - \ln(1.7183) \approx -0.331 - 0.541 = -0.872$
  • So $k \approx 0.920 - 0.872 = 0.048$ (a small positive constant)

内容的提问来源于stack exchange，提问作者Mr Pie